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140. 单词拆分 II

English Version

题目描述

给定一个字符串 s 和一个字符串字典 wordDict ，在字符串 s 中增加空格来构建一个句子，使得句子中所有的单词都在词典中。以任意顺序 返回所有这些可能的句子。

注意：词典中的同一个单词可能在分段中被重复使用多次。

 

示例 1：

输入:s = "catsanddog", wordDict = ["cat","cats","and","sand","dog"]
输出:["cats and dog","cat sand dog"]

示例 2：

输入:s = "pineapplepenapple", wordDict = ["apple","pen","applepen","pine","pineapple"]
输出:["pine apple pen apple","pineapple pen apple","pine applepen apple"]
解释: 注意你可以重复使用字典中的单词。

示例 3：

输入:s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
输出:[]

 

提示：

• 1 <= s.length <= 20
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 10
• s 和 wordDict[i] 仅有小写英文字母组成
• wordDict 中所有字符串都 不同

解法

方法一：前缀树 + DFS

Python3

class Trie:
    def __init__(self):
        self.children = [None] * 26
        self.is_end = False

    def insert(self, word):
        node = self
        for c in word:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                node.children[idx] = Trie()
            node = node.children[idx]
        node.is_end = True

    def search(self, word):
        node = self
        for c in word:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                return False
            node = node.children[idx]
        return node.is_end


class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
        def dfs(s):
            if not s:
                return [[]]
            res = []
            for i in range(1, len(s) + 1):
                if trie.search(s[:i]):
                    for v in dfs(s[i:]):
                        res.append([s[:i]] + v)
            return res

        trie = Trie()
        for w in wordDict:
            trie.insert(w)
        ans = dfs(s)
        return [' '.join(v) for v in ans]

Java

class Trie {
    Trie[] children = new Trie[26];
    boolean isEnd;

    void insert(String word) {
        Trie node = this;
        for (char c : word.toCharArray()) {
            c -= 'a';
            if (node.children[c] == null) {
                node.children[c] = new Trie();
            }
            node = node.children[c];
        }
        node.isEnd = true;
    }

    boolean search(String word) {
        Trie node = this;
        for (char c : word.toCharArray()) {
            c -= 'a';
            if (node.children[c] == null) {
                return false;
            }
            node = node.children[c];
        }
        return node.isEnd;
    }
}

class Solution {
    private Trie trie = new Trie();

    public List<String> wordBreak(String s, List<String> wordDict) {
        for (String w : wordDict) {
            trie.insert(w);
        }
        List<List<String>> res = dfs(s);
        return res.stream().map(e -> String.join(" ", e)).collect(Collectors.toList());
    }

    private List<List<String>> dfs(String s) {
        List<List<String>> res = new ArrayList<>();
        if ("".equals(s)) {
            res.add(new ArrayList<>());
            return res;
        }
        for (int i = 1; i <= s.length(); ++i) {
            if (trie.search(s.substring(0, i))) {
                for (List<String> v : dfs(s.substring(i))) {
                    v.add(0, s.substring(0, i));
                    res.add(v);
                }
            }
        }
        return res;
    }
}

Go

type Trie struct {
	children [26]*Trie
	isEnd    bool
}

func newTrie() *Trie {
	return &Trie{}
}
func (this *Trie) insert(word string) {
	node := this
	for _, c := range word {
		c -= 'a'
		if node.children[c] == nil {
			node.children[c] = newTrie()
		}
		node = node.children[c]
	}
	node.isEnd = true
}
func (this *Trie) search(word string) bool {
	node := this
	for _, c := range word {
		c -= 'a'
		if node.children[c] == nil {
			return false
		}
		node = node.children[c]
	}
	return node.isEnd
}

func wordBreak(s string, wordDict []string) []string {
	trie := newTrie()
	for _, w := range wordDict {
		trie.insert(w)
	}
	var dfs func(string) [][]string
	dfs = func(s string) [][]string {
		res := [][]string{}
		if len(s) == 0 {
			res = append(res, []string{})
			return res
		}
		for i := 1; i <= len(s); i++ {
			if trie.search(s[:i]) {
				for _, v := range dfs(s[i:]) {
					v = append([]string{s[:i]}, v...)
					res = append(res, v)
				}
			}
		}
		return res
	}
	res := dfs(s)
	ans := []string{}
	for _, v := range res {
		ans = append(ans, strings.Join(v, " "))
	}
	return ans
}

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